解答
3sin2(a)cos2(a)=(cot(a))2
解答
a=2π+2πn,a=23π+2πn
+1
度数
a=90∘+360∘n,a=270∘+360∘n求解步骤
3⋅sin2(a)cos2(a)=(cot(a))2
两边减去 (cot(a))2sin2(a)3cos2(a)−cot2(a)=0
化简 sin2(a)3cos2(a)−cot2(a):sin2(a)3cos2(a)−cot2(a)sin2(a)
sin2(a)3cos2(a)−cot2(a)
将项转换为分式: cot2(a)=sin2(a)cot2(a)sin2(a)=sin2(a)3cos2(a)−sin2(a)cot2(a)sin2(a)
因为分母相等,所以合并分式: ca±cb=ca±b=sin2(a)3cos2(a)−cot2(a)sin2(a)
sin2(a)3cos2(a)−cot2(a)sin2(a)=0
g(x)f(x)=0⇒f(x)=03cos2(a)−cot2(a)sin2(a)=0
分解 3cos2(a)−cot2(a)sin2(a):(3cos(a)+cot(a)sin(a))(3cos(a)−cot(a)sin(a))
3cos2(a)−cot2(a)sin2(a)
将 cot2(a)sin2(a) 改写为 (cot(a)sin(a))2
cot2(a)sin2(a)
使用指数法则: ambm=(ab)mcot2(a)sin2(a)=(cot(a)sin(a))2=(cot(a)sin(a))2
=3cos2(a)−(cot(a)sin(a))2
将 3cos2(a)−(cot(a)sin(a))2 改写为 (3cos(a))2−(cot(a)sin(a))2
3cos2(a)−(cot(a)sin(a))2
使用根式运算法则: a=(a)23=(3)2=(3)2cos2(a)−(cot(a)sin(a))2
使用指数法则: ambm=(ab)m(3)2cos2(a)=(3cos(a))2=(3cos(a))2−(cot(a)sin(a))2
=(3cos(a))2−(cot(a)sin(a))2
使用平方差公式: x2−y2=(x+y)(x−y)(3cos(a))2−(cot(a)sin(a))2=(3cos(a)+cot(a)sin(a))(3cos(a)−cot(a)sin(a))=(3cos(a)+cot(a)sin(a))(3cos(a)−cot(a)sin(a))
(3cos(a)+cot(a)sin(a))(3cos(a)−cot(a)sin(a))=0
分别求解每个部分3cos(a)+cot(a)sin(a)=0or3cos(a)−cot(a)sin(a)=0
3cos(a)+cot(a)sin(a)=0:a=2π+2πn,a=23π+2πn
3cos(a)+cot(a)sin(a)=0
使用三角恒等式改写
cos(a)3+cot(a)sin(a)
使用基本三角恒等式: cot(x)=sin(x)cos(x)=cos(a)3+sin(a)cos(a)sin(a)
sin(a)cos(a)sin(a)=cos(a)
sin(a)cos(a)sin(a)
分式相乘: a⋅cb=ca⋅b=sin(a)cos(a)sin(a)
约分:sin(a)=cos(a)
=3cos(a)+cos(a)
cos(a)+cos(a)3=0
用替代法求解
cos(a)+cos(a)3=0
令:cos(a)=uu+u3=0
u+u3=0:u=0
u+u3=0
分解 u+u3:(1+3)u
u+u3
因式分解出通项 u=u(1+3)
(1+3)u=0
两边除以 1+3
(1+3)u=0
两边除以 1+31+3(1+3)u=1+30
化简u=0
u=0
u=cos(a)代回cos(a)=0
cos(a)=0
cos(a)=0:a=2π+2πn,a=23π+2πn
cos(a)=0
cos(a)=0的通解
cos(x) 周期表(周期为 2πn):
x06π4π3π2π32π43π65πcos(x)12322210−21−22−23xπ67π45π34π23π35π47π611πcos(x)−1−23−22−210212223
a=2π+2πn,a=23π+2πn
a=2π+2πn,a=23π+2πn
合并所有解a=2π+2πn,a=23π+2πn
3cos(a)−cot(a)sin(a)=0:a=2π+2πn,a=23π+2πn
3cos(a)−cot(a)sin(a)=0
使用三角恒等式改写
cos(a)3−cot(a)sin(a)
使用基本三角恒等式: cot(x)=sin(x)cos(x)=cos(a)3−sin(a)cos(a)sin(a)
sin(a)cos(a)sin(a)=cos(a)
sin(a)cos(a)sin(a)
分式相乘: a⋅cb=ca⋅b=sin(a)cos(a)sin(a)
约分:sin(a)=cos(a)
=3cos(a)−cos(a)
−cos(a)+cos(a)3=0
用替代法求解
−cos(a)+cos(a)3=0
令:cos(a)=u−u+u3=0
−u+u3=0:u=0
−u+u3=0
分解 −u+u3:(−1+3)u
−u+u3
因式分解出通项 u=u(−1+3)
(−1+3)u=0
两边除以 −1+3
(−1+3)u=0
两边除以 −1+3−1+3(−1+3)u=−1+30
化简u=0
u=0
u=cos(a)代回cos(a)=0
cos(a)=0
cos(a)=0:a=2π+2πn,a=23π+2πn
cos(a)=0
cos(a)=0的通解
cos(x) 周期表(周期为 2πn):
x06π4π3π2π32π43π65πcos(x)12322210−21−22−23xπ67π45π34π23π35π47π611πcos(x)−1−23−22−210212223
a=2π+2πn,a=23π+2πn
a=2π+2πn,a=23π+2πn
合并所有解a=2π+2πn,a=23π+2πn
合并所有解a=2π+2πn,a=23π+2πn